IMO Team Selection Test 2

We first show that $f(f(m))=m$ for all $m$. We start by substituting $n=1$. This yields $f(f(m)f(1))=m$. Then on the one hand $f$ is injective, since if $f(m)=f(n)$ then it immediately follows that $m=f(f(m)f(1))=f(f(n)f(1))=n$. On the other hand, $f$ is surjective, since for every $m$ there is an $a$ such that $f(a)=m$, namely $a=f(m)f(1)$. Hence $f$ is bijective. Now we substitute $m=n$, which yields $f(f(m{)}^2)=m^2=f(f(m^2)f(1))$. Because of the injectivity of $f$, it follows $$f(m{)}^2=f(m^2)f(1).$$ Now choose $m_1$ for which $f(m_1)=1$. (It follows from our proof of the surjectivity of $f$ that we can choose $m_1=f(1{)}^2$ for this, but that is not important.) If we substitute $m=m_1$ in the above we obtain $1=f(m_1^2)f(1)$. In particular, $f(1)$ is a divisor of $1$, and since it must be a natural number we conclude that $f(1)=1$. Substituting $n=1$, now yields $f(f(m))=m$. Now we can substitute $f(m)$ and $f(n)$ for $m$ and $n$ respectively, so that we obtain the more classical equation $$f(mn)=f(f(f(m))f(f(n)))=f(m)f(n).$$ It follows that $f$ is determined by the values $f(p)$ with $p$ prime. Let $p$ be a prime number, and let $m$ and $n$ be natural numbers such that $f(p)=mn$. Then it follows that $f(m)f(n)=f(mn)=f(f(p))=p$. Since $f(m)$ and $f(n)$ are natural numbers, it must hold that $f(m)=1$ or $f(n)=1$. Because of injectivity, the only $m$ for which $f(m)=1$ is the number $m=1$. So either $m=1$ or $n=1$. We conclude that $f(p)$ is a prime number. Now suppose that $p\nmid 2024$ holds. In particular, we have $\gcd (p,2024)=1$, so $p$ has a multiplicative inverse modulo $2024$. Let $b$ be a natural number in the residue class of this inverse, i.e. $bp\equiv 1(\mathrm{mod}2024)$. Since $bp\ge p>1$, it follows from the second equation that $f(bp)=bp$, but also that $f(bp+2024p)=bp+2024p$. However, it follows from the previous paragraph that $f(bp)=f(b)f(p)$ and $f((b+2024)p)=f(b+2024)f(p)$. Therefore $f(p)$ is a divisor of both $bp$ and $bp+2024p$. Now we calculate $$\gcd (bp,bp+2024p)=\gcd (bp,2024p)=\gcd (b,2024)p=p,$$ since $b$ is the multiplicative inverse of $p$ modulo $2024$. Because of injectivity, we know that $f(p)\ne 1$, so $f(p)\mid p$ implies that $f(p)=p$. Suppose for a prime divisor $p$ of $2024$ that $f(p)=q$, with $q\ne p$ a different prime number. Then it follows that $f(q)=f(f(p))=p$. So all prime numbers not mapped to themselves by $f$ form pairs $(p,q)$ with $f(p)=q$ and $f(q)=p$. Since $2024=2^3\cdot 11\cdot 23$, we have four possible cases: $f(p)=p$ for all primes $p$. So $f$ is the identity, $f(2)=11$, $f(11)=2$ and $f(p)=p$ for all $p\nmid 2,11$, $f(2)=23$, $f(23)=2$ and $f(p)=p$ for all $p\nmid 2,23$, $f(11)=23$, $f(23)=11$ and $f(p)=p$ for all $p\nmid 11,23$. It is easy to check that all four of these cases actually lead to a function that satisfies the given conditions. $\square$